ORBIT SIZES and the DIHEDRAL GROUP of ORDER EIGHT By

ORBIT SIZES AND THE DIHEDRAL GROUP OF ORDER EIGHT

Nathan A. Jones, B.S.

A thesis submitted to the Graduate College of Texas State University in partial fulﬁllment of the requirements for the degree of Master of Science with a Major in Mathematics May 2018

Committee Members:

Thomas Keller, Chair

Anton Dochtermann

Yong Yang COPYRIGHT by Nathan A. Jones 2018 FAIR USE AND AUTHOR’S PERMISSION STATEMENT

Fair Use

This work is protected by the Copyright Laws of the United States (Public Law 94- 553, section 107). Consistent with fair use as deﬁned in the Copyright Laws, brief quotations from this material are allowed with proper acknowledgment. Use of this material for ﬁnancial gain without the author’s express written permission is not allowed.

Duplication Permission

As the copyright holder of this work I, Nathan A. Jones, authorize duplication of this work, in whole or in part, for educational or scholarly purposes only ACKNOWLEDGMENTS

I would like to thank Dr. Thomas Keller for being my advisor for this thesis. From the start Dr. Keller helped find a great problem to tackle. His work inspired new ways of thinking. During the time we spent together each week he had unpar- alleled patience and encouragement. In addition he taught my first year Algebra where I got to learn the real beauty and elegance of finite group theory. I would also like to thank my committee members, Dr. Yong Yang, Dr. An- ton Dochtermann. Dr. Yang has always been incredibly friendly and encouraging in our encounters. Dr. Dochtermann was gracious and helpful in learning about and preparing this thesis. Finally, I would like to thank my dad who told me “there is no money in archeology, you’ll starve”, starting my journey of mathematical curiosity.

iv TABLE OF CONTENTS

Page

ACKNOWLEDGEMENTS...... iv

CHAPTER

I. INTRODUCTION...... 1

Deﬁnitions...... 1 Useful Theorems and Lemmas...... 6 II. EXAMPLES AND STATEMENT OF RESULT...... 9

III. PROOF OF MAIN RESULT...... 11

A Reduction to Nilpotent Groups...... 11

A Reduction to p-groups...... 12

The Case Where V is Quasiprimitive...... 15

The Case Where V is Quasiprimitive...... 16

The Case Where D′ < G′ ...... 16

The case where D is nonabelian...... 18

The Case Where D is Abelian...... 19

The Case Where D has Exactly One Orbit Size M...... 19

The Case Where D has Exactly Two Orbits of Size M...... 20

The Case where D′ = G′ ...... 22 IV. The Open Case...... 36

REFERENCES...... 37 I. INTRODUCTION

Groups form one of the most basic structure available in abstract algebra, finding uses in a variety of areas in mathematics. In number theory the main object of study is the set of integers, a group. The solutions to polynomial equations give rise to a group. Well known results such as Euler’s Theorem, Diophantine equations, and solvability of polynomial equations have all benefited from the under- standing of groups and their structure [3, p.14]. A natural tool for studying groups is a group action on a set. Group actions provide a convenient way to study abstract groups. Seeing where a group can send a particular element of a set gives you a sense of that element’s orbit. One question that arises is to study the sizes of these orbits. In this thesis we will be concerned with a lower bound for the largest orbit size. We will primarily concern ourselves with group actions where G is a finite nonabelian group acting on a finite faithful

irreducible G-module V . It is known that SGS~SG′S, where G′ is the commutator subgroup, serves as a lower bound for the largest orbit size of G acting on V [8]. It is even known that this inequality is either strict, or equal and G is abelian, or there are at least two orbits of the largest orbit size in the action of G on V when G is nonabelian. The next natural step is to consider what happens if there are exactly two orbits of size SGS~SG′S in the action of G on V . In the cited paper the authors conjecture that this can only happen when the group is the dihedral group of order eight, while V = V (2, 3) is a two dimensional vector space over a ﬁeld of 3 elements. In this thesis we will show that a slightly weaker version of this conjecture is in fact true. Deﬁnitions

In this section we will formalize the terms used throughout this thesis. We

1 will also clarify notation which may vary in different sources. The definitions used in this paper have been taken from [6, 10] and the notation is consistent with litera- ture in finite group theory. This thesis will assume the reader is familiar with the basic concept of a group. A quick review of the properties of a group include a set G which is closed under a binary operation which is associative, contains an identity, and has in- verses. In this thesis all groups will be finite. That is to say the size of set G, denoted SGS, is not infinite. An example of a group is the dihedral group on eight elements, denoted D8. This group consists of the four rotational and four ’flip’ sym-

2 3 2 3 metries of a square. These are represented by D8 = {1, r, r , r , s, sr, sr , sr }, where r can be considered a 90 degree rotation of a square, and s can be considered a ﬂip over a ﬁxed axis of symmetry of a square. This produces the multiplication rule

sr = r−1s. The ﬁrst structures on a group G we will consider are smaller groups in- side G called subgroups. Subgroups provide a way to get an idea of the structure of larger groups. In this paper subgroups will be used to allow for induction on the

2 3 group. In the case of D8 we can look at only the set of rotations {1, r, r , r } and

see this forms a subgroup of D8. We will deﬁne some more special subgroups.

Deﬁnition 1. A normal subgroup is a subgroup H ≤ G such that for all g ∈ G we have g−1Hg = H. This is denoted by H ⊴ G, and H ⊲ G if and only if H ⊴ G and H ≠ G.

Notice that a normal subgroup is a group that is invariant under conjugation by elements in G. These groups are also known to be characterized as the kernel of some group homomorphism.

Deﬁnition 2. A group G is solvable if there exists a chain of subgroups 1 = H0 ⊲

2 −1 H1 ⊲ ... ⊲ Hn = G such that Hi+1~Hi is abelian for i = 0, ..., n .

Notice this means that solvable groups are groups that are constructed by extensions of abelian groups. Solvable groups will be the focus of this paper.

Deﬁnition 3. The Frattini subgroup,Φ(G) is the intersection of all maximal proper subgroups of G.

The Frattini subgroup always exists in ﬁnite groups and possesses many useful properties. For example, the Frattini subgroup is always normal. In the case of

2 2 2 3 2 3 D8 we have three maximal subgroups {1, s, r , sr } , {1, r, r , r }, and {1, rs, r , sr }.

2 Therefore Φ(D8) = {1, r }. One property of the Frattini subgroup we will beneﬁt from is that it is nilpotent.

Deﬁnition 4. Let G be a ﬁnite group and p a prime. A Sylow p-subgroup of G is a subgroup P ≤ G such that SP S = pa is the full power of p dividing SGS. The set of

all Sylow p-subgroups of G is denoted Sylp(G)

Deﬁnition 5. Let G be ﬁnite and solvable and let π be any set of prime numbers.

Hall’s Theorem guarantees a subgroup H ≤ G with order divisible only by primes in π with SGS~SHS divisible by none of these primes. A subgroup H ≤ G satisfying these conditions is called a Hall π-subgroup of G.

The Sylow and Hall subgroups will provide us a natural way to ‘split-up’ groups into two subgroups using a free product.

Deﬁnition 6. A group G is called nilpotent if there exists a chain of subgroups

1 = G0 ⊲ G1 ⊲ ... ⊲ Gn = G such that Gi+1~Gi ≤ Z(G~Gi). Where Z(G~Gi) = {g ∈

G~GiSgx = xg for all x ∈ G~Gi}.

3 This formal deﬁnition can be replaced by another property that is equiva- lent to being nilpotent. A group G is nilpotent if and only if it can be written as the direct product of its Sylow subgroups for all primes p dividing SGS. Notice that because D8 is a ﬁnite p-group that it must be a nilpotent group.

Deﬁnition 7. The Fitting subgroup of G, written as F (G) is the unique largest normal nilpotent subgroup of G.

In the example of D8, we would have F (D8) = D8, because D8 is nilpotent. The Fitting subgroup will appear in relation to Gasch¨utz’Theorem which we will state in the next section.

Deﬁnition 8. Let H1 and H2 be subgroups of G. We deﬁne the commutator of these groups to be

−1 −1 [H1,H2] = ⟨h1 h2 h1h2Sh1 ∈ H1, h2 ∈ H2⟩.

The commutator subgroup of G is the group [G, G] and denoted by G′.

The commutator subgroup is the smallest normal subgroup such that the quotient group of the group by its commutator is abelian. It turns out that SG~G′S has a relation to the largest orbit size of a group action. In the case of D8 the commutator subgroup is {1, r2}.

Deﬁnition 9. Let X be a set and G be a group. We say that G acts on X if for

h every x ∈ X and g ∈ G there exists an element xg ∈ X such that x1 = x and xg = xgh for all g, h ∈ G. If G acts on X, we call this a right group action.

The group action is an abstraction that appears naturally among many ob- jects in mathematics. Therefore they are worth studying in their own right. We will discuss a few properties of group actions next section.

4 Deﬁnition 10. Let G act on the set X and x ∈ X. The orbit containing x of this group action is the set {xgSg ∈ G}

An orbit of an element x can be thought of informally as elements in the set that x can be taken to by an element in the group.

Deﬁnition 11. A group action is faithful if there is no g ∈ G where g ≠ 1 such that xg = x for all x ∈ X.

One will notice that a faithful action induces an injection from G to the symmetric group on X.

Deﬁnition 12. A group action of G on X is said to be transitive if for every two elements x, y ∈ X, there exists g ∈ G with xg = y. If this g is unique we say that the action is regular.

Deﬁnition 13. Let G be a group and V vector space over a ﬁeld. Let G act on V such that (a + b)g = ag + bg for all a, b ∈ V and g ∈ G. We call V a G-module.

Deﬁnition 14. Let D be a subgroup of G and V be a G-module. If we consider only the action of D on V we get a D-module denoted VD.

Deﬁnition 15. Let V,W be G-modules. We say V ≅ W as G-modules if and only if there exists a vector space isomorphism φ ∶ V → W such that φ(vg) = φ(v)g for all v ∈ V and g ∈ G.

Deﬁnition 16. A G-module V is irreducible if V has no proper non-zero G- submodules.

Deﬁnition 17. A G-module V is completely reducible if it can be represented as the direct sum of irreducible G-modules.

5 Deﬁnition 18. Let V be a completely reducible G-module. Then V = V1 ⊕V2 ⊕...⊕

Vn for irreducible G-modules Vi. One can write V = W1 ⊕ ... ⊕ Wm for some m ≤ n such that Vi,Vj ≤ Wk for some i, j ∈ {1, ..., n} and some k ∈ {1, ..., m} if and only if

Vi ≅ Vj (as G-modules). Then the Wi are called the homogeneous components of V .

Deﬁnition 19. An irreducible G-module V is called imprimitive if V can be written as V = V1 ⊕ ... ⊕ Vn for n > 1 subspaces Vi that are permuted transitively by G. We say that V is primitive if V is not imprimitive. V is called quasi-primitive if VN is homogeneous for all N ⊲ G (where VN denotes V viewed as an N-module).

It is known that quasiprimitive is a weaker condition than primitive, that is primitive implies quasiprimitive. However the reverse implication is not true. This concludes the formal deﬁnitions that will be required for the main result.

Useful Theorems and Lemmas

The following results are all known and appear in other papers relating to group theory. Their original sources have been indicated for proofs of the following results. The first lemma will come in handy when we examine abelian subgroups in our main theorem. Lemma 1. [8, Lemma 2.2] Let A be an abelian finite group and let V be a finite faithful completely reducible A-module. It is well-known that A has a regular orbit on V . Write V = V1 ⊕ ... ⊕ Vn for irreducible A-modules Vi. Suppose that A has exactly one regular orbit on V . Then A~CA(Vi) is cyclic of order SViS − 1 for all i n n and A ≅ ⨉i=1 A~CA(Vi) is of order ∏i=1(SViS − 1).

The following theorem is of key importance in our paper. It is the main re-

6 sult of [8] and will be used throughout the paper. In particular this paper is concerned with expanding the third case, however the ﬁrst two cases will also be used to show our result. The following theorem will reference modules of mixed characteristic. Mixed characteristic means that if V = V1 ⊕ ...Vn then Vi and Vj need not have the same characteristic for all i, j ∈ {1, . . . , n}

Theorem 1. [8, Theorem 2.3] Let G be a ﬁnite solvable group and V a ﬁnite faithful completely reducible G-module, possibly of mixed characteristic. Let M be the largest orbit size in the action of G on V . Then

′ SG~G S ≤ M

More precisely, we have one of the following

1. SG~G′S < M

2. SG~G′S = M and G is abelian; or

3. SG~G′S = M, G is nilpotent, and G has at least two diﬀerent orbits of size M on V .

The following Lemma is well-known and very useful. The reader may recog- nize this as a consequence of the isomorphism theorems.

Lemma 2. [8, Lemma 2.1] Let G be a ﬁnite group and N ⊴ G. Then

′ ′ ′ SG~G S = SG~G NS ⋅ SN ∶ N ∩ G S; and

′ ′ ′ SG ∶ G S divides SG~N ∶ (G~N) S ⋅ SN ∶ N S.

7 Gasch¨utz’Theorem will be used in the main proof to make a reduction to nilpotent groups. This will allow us to narrow the scope of possible cases needed to prove the result.

Theorem 2 (Gasch¨utz’Theorem). [10, p.37] Let G be solvable. Then F (G~Φ(G)) = F (G)~Φ(G) is a completely reducible and faithful G~F (G)-module (possibly of mixed characteristic). Furthermore, G~Φ(G) splits over F (G)~Φ(G).

Lemma 3. Let D8 ≤ GL(2, 3) (invertible 2x2 matrices on the ﬁeld of three elements) act on V (2, 3) = V . Let N ⊲ D8 with N being a Klein-4 group. Then N has exactly one orbit of size four on V . If K is the other Klein-4 subgroup of D8, then the orbit of size 4 from K acting on V is diﬀerent than the orbit of size four of N acting on V .

2 2 3 Proof. Let N = {1, r, s, sr } and K = {1, r , sr, sr }, the two subgroups of D8 isomorphic to the Klein-4 group. Denoting V (2, 3) by (i, j) where i and j are integers such that −1 ≤ i ≤ 1 and −1 ≤ j ≤ 1. Using right multiplication with ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢0 −1⎥ ⎢1 0 ⎥ r ⎢ ⎥ , s ⎢ ⎥ we can calculate the two orbits of size four. When K = ⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢1 0 ⎥ ⎢0 −1⎥ ⎣ ⎦ ⎣ ⎦ acts on V the orbit of size four is {(1, 1), (1, −1), (−1, 1), (−1, −1)} while the orbit of size four in the action of N on V is {(1, 0), (−1, 0), (0, −1), (0, 1)}. This concludes the proof of Lemma 3

8 II. EXAMPLES AND STATEMENT OF RESULT

The following theorem is similar to a conjecture in [8] before becoming the topic of this thesis. It will expand upon the third case of Theorem 1 in this paper

by considering the case where G is nonabelian, SGS is not divisible by a Mersenne prime and has exactly two orbits of size M on V and V is irreducible. Our claim is that this can only happen if and only if G = D8 and V = V (2, 3), the vector space of dimension two over the ﬁeld of three elements. These orbits can be calculated by the reader using the following matricies for r and s. The operation will be right multiplication of matrices on the elements of V . We can generate D8 by using the following matrices to generate a subset of the general linear group GL(2, 3).

⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢0 −1⎥ ⎢1 0 ⎥ r ⎢ ⎥ , s ⎢ ⎥ = ⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢1 0 ⎥ ⎢0 −1⎥ ⎣ ⎦ ⎣ ⎦ The orbits can also be represented in the following picture. Imagine arrang- ing these points in a 3 by 3 grid as they appear in the familiar xy-axis. The two orbits can be seen with the connecting lines, both of size 4, while (0, 0) remains as a ﬁxed point. Notice we see two squares, the symmetries of which are described by

D8.

9 (−1, 1) (0, 1) (1, 1)

(−1, 0) (0, 0) (1, 0)

(−1, −1) (0, −1) (1, −1)

′ ′ If we recall the size of D8 and the size of D8 we see that SD8 ∶ D8S = 4 and this action satisﬁes conditions of the next theorem.

Theorem 3. Let G be a ﬁnite nonabelian solvable group and V a ﬁnite faithful irreducible G-module Suppose that M = SG~G′S is the largest orbit size of G on V ,SGS is not divisible by a Mersenne prime, and that there are exactly two orbits of size

M on V . Then G is dihedral of order 8, and V = V (2, 3).

We note that the original conjecture did not contain the added hypothesis that SGS is not divisible by Mersenne prime. We still believe that the conjecture holds without this condition, but have not found a way to dispense of the added hypothesis in one case of our proof. Further work will be done in order to remove this extra hypothesis.

10 III. PROOF OF MAIN RESULT

Proof. In this proof we will use induction on the size of our group SGS several times. We begin the proof by making a reduction and showing that if G is not a nilpotent group it will not satisfy the conditions of our theorem. If G is not nilpotent then the proof of Theorem 2.3 in [8] shows that the largest orbit size of G on V is less than M. Here we reproduce the argument for the convenience of the reader.

A Reduction to Nilpotent Groups

Suppose G is not nilpotent. Write F = F (G) for the Fitting subgroup and Φ = Φ(G) for the Frattini subgroup of G. As G is not nilpotent, we have F < G, and since F is normal in G, V is completely reducible as an F -module. Hence by Theorem 2.0.2 applied to the action of F on V , we have that

′ SF ~F S ≤ M. (1)

Moreover, by Gasch¨utz’Theorem it is well-known that F ~Φ is a faithful, completely reducible G~F −module (possibly of mixed characteristic). We now write F ~Φ =

′ W1 ⊕ W2, where W1 = (F ∩ G )Φ~Φ and W2 is G-invariant complement of W1 in F ~Φ. Hence

′ W2 ≅ (F ~Φ)~W1 ≅ F ~(F ∩ G Φ) as G-modules.

We now claim that G~F acts trivially on W2. For this it suﬃces to show that G acts trivially on F ~F ∩ G′Φ. So let g ∈ G and x ∈ F . Then

11 ′ g g ′ ′ ′ (x(F ∩ G )Φ) = x (F ∩ G )Φ = x[x, g](F ∩ G )Φ = x(F ∩ G )Φ,

the last equality being true since [x, g] ∈ F ∩ G′. This proves our claim.

Now as G~F is faithful on F ~Φ, but acts trivially on W2, we see that W1 is a faithful completely reducible G~F -module (of possibly mixed characteristic). Thus by Theorem 2.0.2 we conclude that

′ ′ SG ∶ FG S = S(G~F ) ∶ (G~F ) S < SW1S. (2)

So altogether with (3.1) and (3.2) we conclude that

′ ′ ′ ′ ′ ′ ′ ′ SG~G S = SG ∶ FG SSFG ~G S < SW1S ⋅ SFG ~G S = S(F ∩ G )Φ~ΦSSF ∶ (F ∩ G )S

′ ′ ′ = S(F ∩ G )~(F ∩ G ∩ Φ)SSF ∶ (F ∩ G )S

′ ′ ′ ≤ S(F ∩ G ) ∶ F S ⋅ SF ∶ (F ∩ G )S

′ = SF ~F S ≤ M

This gives us SG~G′S < M contradicting our hypothesis. Therefore we know that G must be nilpotent. This concludes the argument from [8]. A Reduction to p-groups

We can further make a reduction to p-groups. Assume that SGS is divisible

by at least two distinct primes, one of which we will call p. Let P ∈ Sylp(G) and

H ∈ Hallp′ (G). It is known that because G is nilpotent, P ⊲ G [6, p.89]. Moreover we can write G = P × H [6, Theorem 8.11]. V is a ﬁnite G-module over a ﬁeld, call

12 it K. By [11, Lemma 10] there exists a ﬁeld extension L of K such that if U is an irreducible summand of V viewed as an LG-module, then the permutation actions of G on V and U are permutation isomorphic. We may consider the action of G on U instead. With relabeling we can assume that V is absolutely irreducible. By [1,

(3.6)] we may assume V = X1 ⊗ X2, where X1 is a faithful irreducible P -module and

X2 is a faithful irreducible H-module. By Theorem 2.0.2 we can choose x1 ∈ X1 and

x2 ∈ X2 such that ′ P SP ~P S ≤ Sx1 S,

′ H SH~H S ≤ Sx2 S.

P H Clearly we may assume that Sx1 S is the largest orbit size of P on X2, and Sx2 S is the

largest orbit size of H on X2. Using [9, Lemma 3.3] if g ∈ P and h ∈ H such that gh ∈ CG(x1 ⊗ x2) then x1g = αx1 and x2h = βx2 where α, β are scalars in the ﬁeld with αβ = 1. Now g and h have coprime orders so we have that α = β = 1 which gives

CG(x1 ⊗ x2) = CP (x1) × CH (x2).

This gives the following

′ G SG~G S = M ≥ S(x1 ⊗ x2) S = SG ∶ Cp(x1) × CH (x2)S (3)

= SP ∶ CP (x1)SSH ∶ CH (x2)S

P H ′ ′ ′ = Sx1 SSx2 S ≥ SP ~P SSH~H S = SG~G S.

P ′ This shows that we have equality everywhere in 3, and so Sx1 S = SP ~P S, and

H ′ ′ ′ Sx2 S = SH~H S. Therefore SP ~P S is the largest orbit size of P on X1 and SH~H S is the largest orbit size of H on X2.

13 ′ Suppose that P and H have exactly one maximal orbit of size M1 = SP ~P S

′ and M2 = SH~H S on X1 and X2 respectively. By Theorem 1 P and H would be abelian groups. Then G = P × H would be abelian, a contradiction. Therefore, one of the following holds, P has at least two orbits of size M1 on X1, or H has at least two orbits size M2 on X2.

We will now show that P has no more than two orbits of size M1 on X1 and

H has no more than two orbits of size M2 on X2. The argument will be the same in both cases. We assume P has three orbits of maximal size on X1. Let y1, y2 ∈ X1 be the representatives of the remaining two orbits. Then as in (3.3)

G G G S(x1 ⊗ x2) S = S(y1 ⊗ x2) S = S(y2 ⊗ x2) S = M, making three orbits of size M when G acts on V contradicting our hypothesis. Us- ing a similar argument we see that H cannot have more than two orbits of size M2 on X2.

We now will show that either P cannot have two orbits of size M1 on X1 or

H cannot have two orbits of size M2 on X2. If P and H both have two orbits of maximal size we have y1 ∈ X1 as a second representative of an orbit size M1 and y2 ∈ X2 a second representative of an orbit of size M2 of P and H, respectively. Whenever it is assumed that P and H have a second orbit we will call upon these elements to represent the second orbit. Then, again proceeding as in (3.3), we see that

′ G G G G SG~G S = S(x1 ⊗ x2) S = S(x1 ⊗ y2) S = S(y1 ⊗ x2) S = S(y1 ⊗ y2) S = M creating four orbits of size M in the action of G on V , a contradiction. Therefore P has exactly two orbits of maximal size on X1 or H has exactly two orbits of maxi-

14 mal size on X2 but not both. Next we consider what happens in these two cases.

′ First we consider the case where P has two orbits of size SP ~P S on X1. H

′ G must have exactly one orbit of size SH~H S on X2. Notice that (x1 ⊗ x2) and (y1 ⊗

G x2) are two distinct orbits of G of size M. By part 2 of Theorem 1 H must be

′ H abelian. Thus SH~H S = SHS = Sx2 S, by Lemma 1 H ≅ H~CH (X2) and H is a cyclic

group of order SX2S − 1. We know that P is not abelian, since if it were, then G =

P × H would be abelian. Therefore we can use induction, P = D8 and X1 = V (2, 3).

n This makes char(X1) = 3 so char(X2) = 3. Therefore SHS = 3 − 1 for some n ∈ N

making SHS even. This contradicts that P ∈ Syl2(G). Therefore P cannot have two

′ orbits size SP ~P S on X2.

′ We now consider the case that H has two orbits of size SH~H S on X2. Then

′ P has one orbit of size SP ~P S on X1. By Theorem 1 part 2 P is an abelian group

and by Lemma 1 P is cyclic of order SX1S − 1. We can see that H is not abelian, if

it were G = H × P would be abelian. Using induction we see that H = D8 and X2 =

n V (2, 3). This means SHS = 8 and char(X1) =char(X2) = 3 and SP S = SX1S − 1 = 3 − 1, for some n ∈ N. Therefore SP S is even contradicting gcd(SHS, SGS) = 1. This shows that G must be a p-group. The Case Where V is Quasiprimitive

Consider the case where V is quasiprimitive. Using the proof of Theorem

3.3 in [10] we know G = S × T where T is cyclic of odd order and S a 2-group. There also exists a U ⊲ G where U is cyclic, SG ∶ US ≤ 2 and U has a regular orbit on V . This gives the inequality M ≥ SUS ≥ SGS~2. Because G is a nonabelian p-group we have ST S = 1 and G = S, and p = 2. By Corollary 1.3 in [10] G is cyclic, quaternion, dihedral, or semi-dihedral and G ≇ D8. G is not abelian so SGS > 4 and 8SSGS. It is also known that the derived subgroup of the quaternion,

15 dihedral, and semi-dihedral 2-groups have index 4 [7, proof of Theorem A]. That is

SG~G′S = 4 ≤ SGS~2 ≤ M. Here we have equality so SG′S = 2, SGS = 8. This makes G the quaternion group. It is known that the quaternion group has a regular orbit on V

[10, Lemma 4.2(a)] contradicting M = 4. Therefore V can not be quasiprimitive. The Case Where V is Quasiprimitive

From now on we may assume that V is not quasiprimitive. In particular, V is imprimitive. By Corollary 0.3 in [10] there exists a D ⊴ G with SG ∶ DS = p where p is prime and VD = V1⊕...⊕Vp for irreducible D-modules Vi of VD. We will consider two cases, D′ < G′ and D′ = G′. The Case Where D′ < G′

For the ﬁrst case suppose that D′ < G′, that is, pSD′S ≤ SG′S. Using Theorem 1 we have the following inequality

′ SDS pSDS SGS M ≥ SD ∶ D S = = ≥ = M. SD′S pSD′S SG′S

′ p Therefore SD ∶ D S = M. Because ∩i=1CG(Vi) = 1 we have that D is isomorphic to a subgroup of D~CD(V1) × ... × D~CD(Vp) [8, equation (4)]. We will use the symbol H ⪅ G to denote that H is isomorphic to a subgroup of G. Then

p D ⪅ D~CD(V1) × ... × D~CD(Vp) = i=1D~CD(Vi) =∶ T. (4)

Notice that the above equation tells us that if D~CD(Vi) is abelian for any i =

1, . . . , p then D~CD(Vi) is abelian for all i = 1, . . . , p and D must be abelian. We will use this fact several times in the following arguments.

From Theorem 1, having SD ∶ D′S = M shows that D is abelian or has exactly

two orbits of size M on VD. We know D cannot have more than two orbits of size

16 M on VD, otherwise G would have more than two orbits of size M on V or an orbit larger than size M.

Recall that VD = V1 ⊕ V2 ⊕ ... ⊕ Vp, with Vi irreducible faithful D-modules. p Let Wi = ⊕j=1,j≠iVj. Write M1 for the largest orbit of size D on V1 and M2 for the

largest orbit of size CD(V1) on W1. Also let MD be the largest orbit size of D on V .

Let x ∈ VD be in a largest orbit of D on VD. Write x = x1 + x2 for some x1 ∈ V1 and x2 ∈ W1. Observe that

MD = SD ∶ CD(x)S = SD ∶ CD(x1) ∩ CD(x2)S =

D CD(x1) SD ∶ CD(x1)SSCD(x1) ∶ CD(x1) ∩ CD(x2)S = Sx1 SSx2 S.

D D If Sx1 S < M1, then the same calculation would show that if y1 ∈ V1 with Sy1 S = M1, D then SD ∶ CD(y1 +x2)S > MD, contradicting the deﬁnition of MD. Thus we have Sx1 S =

CD(x1) CD(V1) CD(x1) M1. Moreover, since Sx2 S ≥ Sx2 S. We also can conclude that Sx2 S ≥ M2,

CD(x1) CD(V1) because if Sx2 S < M2, then let y2 ∈ W1 such that Sy2 S = M2, and then

D MD = S(x1 + y2) S = SD ∶ CD(x1) ∩ CD(y2)S = SD ∶ CD(x1)SSCD(x1) ∶ CD(x1) ∩ CD(y2)S

CD(x1) CD(x1) M1Sy2 S = M1M2 > M1Sx2 S = SD ∶ CD(x)S = MD,

a contradiction. Thus altogether we get M ≥ MD ≥ M1M2. Then

D M ≥ Sx S = SD ∶ CD(x)S = SD ∶ CD(x1) ∩ CD(x2)S

CD(x1) CD(W1) = SD ∶ CD(x1)SSCD(x1) ∶ CD(x1) ∩ CD(x2)S = M1Sx2 S ≥ M1Sx2 S = M1M2.

′ ′ ′ By applying Theorem 2 SD ∶ D S divides SD~CD(V1) ∶ D~CD(V1) SSCD(V1)~CD(V1) S

17 and we have

′ ′ ′ M ≥ M1M2 ≥ SD~CD(V1) ∶ (D~CD(V1) )SSCD(V1) ∶ CD(V1) S ≥ SD ∶ D S = M.

′ This shows M1M2 = MD = M. It also follows that M1 = SD~CD(V1) ∶ (D~CD(V1)) S

′ and M2 = SCD(V1) ∶ CD(V1) S. Suppose there exists y1, z1 ∈ V1 where y1 and z1 are representatives of two new orbits of size M1 of D~CD(V1) acting on V1, then arguing as we have previously x1 + y2, y1 + x2 and z1 + x2 are three representatives of orbits of size M of D on VD. This contradicts our hypothesis. So D~CD(V1) has at most two orbits of size M1 on V1. We also know that if D~CD(V1) has exactly one orbit of size M1 on V1 then D~CD(V1) is abelian by Theorem 1, and by (4) D is abelian. Therefore if D is abelian, then D~CD(V1) has exactly two orbits of size M1 on V1. First we will consider the case that D is nonabelian. The case where D is nonabelian.

We know by (4) that D~CD(V1) is not abelian, for the reason listed above.

As argued before CD(V1) must have exactly one orbit of size M2 on W1, otherwise G would have too many orbits of size M. By induction D~CD(V1) = D8, V1 =

V (2, 3), and p = 2 so SW1S = SV1S = SV2S. We have D ⪅ D~CD(V1)×D~CD(V2) ≅ D8 ×D8 making CD(V1) ⪅ D8. If CD(V1) ≅ D8 then CD(V1) has two orbits of size M2 on V2.

Contradicting CD(V1) having exactly one orbit of size M2 on W1 = V2. Therefore

CD(V1) is of order one, two or four. This makes CD(V1) abelian which gives it at least two regular orbits (i.e, two orbits of size M2 = SCD(V1)S) unless CD(V1) is the Klein four-group.

Since CD(V1) has only one orbit of size M2 on V2, we conclude that CD(V1) is the Klein four-group. Since CD(V1) and CD(V2) are conjugate under the action

18 of G, then CD(V2) is also a Klein four-group. Let v1 ∈ V1 be in an orbit of size M1 =

4 of D~CD(V1) on V1, such that v1 is in a regular orbit of CD(V2) on V1 by Lemma

3. Now SCD(v1)S = 8, CD(v1) ∩ CD(V2) = 1, therefore

CD(v1) ≅ CD(v1)~(CD(v1) ∩ CD(V2)) ≅ CD(v1)CD(V2)~CD(V2) ≅ D~CD(V2) ≅ D8 where the third isomorphism comes from the isomorphism theorems. Hence we have CD(v1) ≅ D8. Therefore CD(v1) acts faithfully on V2 and if z1, z2 ∈ V2 are representatives of the the two orbits of size four in the action fo CD(v1) on V2, then v1 + z1 and v1 + z2 are representatives of two orbits of size 16 = MD = M of D on VD.

Now let w1 ∈ V1 be in an orbit of D of size M1 = 4 such that w1 is not in a regular orbit of CD(V2) on V1. Let z3 ∈ V2 be in the (unique) regular orbit of CD(V1) on V2

(so it is of size M2 = 4). Then clearly w1 + z3 is a representative of an orbit of size

16 = M of D on V , and this orbit is diﬀerent from this orbit containing v1 + z1 and v1 + z2. Thus we have found three orbits of size 16 = M of D on V which contradicts our hypothesis. This concludes the case where D is nonabelian. The Case Where D is Abelian.

We will consider two possibilities, that D has exactly one orbit of size M or D has exactly two orbits of size M on V . The Case Where D has Exactly One Orbit Size M.

In the proof of Theorem 1 in [8] it is shown that this results in G = D8 and SV S = 9. The following argument has been reproduced for completeness. Since D is abelian, D has regular orbits, so M = SDS. Hence, D has exactly p one regular orbit on V , therefore by Lemma 1 we have D = i=1CD(Wi). No- tice that G~D cycles these direct factors around, therefore we see that SG ∶ G′S =

19 ′ p pSCD(W1)S (see [5]). Additionally, SG~G S = SDS = SCD(W1)S which implies that

p−1 SCD(W1)S = p. This forces p = 2 and SCD(W1)S = 2. It follows that SDS = 4, D is elementary abelian, SGS = 8, and G is nonabelian. Because M = SDS = 4, G does not have a regular orbit on V , making G the dihedral group of order 8. D is elementary

abelian and has exactly one regular orbit on V so we conclude that SV S = 9. This is

a second veriﬁcation that D8 satisﬁes our hypothesis and ends the case from [8]. The Case Where D has Exactly Two Orbits of Size M.

Assume that D has exactly two regular orbits on V . As before we will de-

note the largest orbit size of the action of D~CD(V1) on V1 as M1, and M2 will de-

note the largest orbit size of CD(V1) acting on W1. Recall that D~CD(V1) has at

most two orbits of size M1 on V1. So the D~CD(Vi) for i = 1, . . . , p are all isomor-

phic and D~CD(Vi) has either one or two orbits of size M1 acting on Vi.

Suppose we have two orbits of size M1 in the action of D~CD(V1) on V1,

then as argued before CD(V1) has exactly one orbit of size M2 on W1. Using Theo-

rem 1 part (2) we see that CD(V1) is abelian. Using induction we see that D~CD(V1) ≅

D8, however the quotient of an abelian groups must be abelian creating a contradiction.

Hence we know that D~CD(V1) has only one orbit of size M1 in the action

of D~CD(V1) on V1 and CD(V1) has exactly two orbits of size M2 on W1. Hence we

have SD~CD(V1)S = SV1S − 1, and D~CD(V1) is cyclic. Now by (3.4) we have D ⪅ p i=1D~CD(Vi) =∶ T and T has exactly one regular orbit on V . Every regular orbit ST S of T on V splits into SDS regular orbits of D. Since D has no more than two orbits ST S ST S of size M = MD(= SDS), we see that SDS ≤ 2. If SDS = 1, then T = D and so CD(V1) ≅ p i=1D~CD(Vi) has only one regular orbit (i.e., of size M2) on W1, a contradiction.

ST S p Therefore we know that SDS = 2. From Lemma 1 we have that ST S = (SViS − 1) ,

20 thus

p k (SV1S − 1) = 2SDS = 2p (5)

for appropriate k. Hence 2 divides the right hand size, so 2p divides the left hand

size, this forces p = 2 and V = V1 ⊕ V2. We wish to show that SCD(V1)S = 2. Observe

that CD(V1) ∩ CD(V2) = 1 and so CD(V1) × CD(V2) = CD(V1)CD(V2) ≤ D. Let

′ g ∈ G − D and (1, a) ∈ CD(V1) × CD(V2), then G contains the element

−1 −1 −1 g −1 ∗ ∗ −1 [(1, a), g] = (1, a) g (1, a)g = (1, a) (1, a) = (1, a )(a , 1) = (a , a )

∗ for suitable a ∈ CD(V1). If there are more than two choices for a ∈ CD(V2) then

′ SGS SGS ′ SG S ≥ 3. On the other hand, SG′S = SDS and SDS = 2 so SG S = SGS~SDS = 2. We conclude that SCD(V1)S ≤ 2. If SCD(V1)S = 1 then D = D~CD(V1), but we know that D has exactly two orbits of size M1 on V1 while D~CD(V1) has only one orbit of size M1 on V1. Therefore we can say SCD(V1)S = 2.

ST S SD~CD(V1)SSD~CD(V2)S We can now determine SDS using (5) we have that SDS = SDS = SDS SDS SDS2 2. This implies 2SDS = 2 2 = 4 , or SDS = 8 and SGS = 16. From Lemma 1

SD~CD(V1)S = SV1S−1 giving us SV1S = 5, i = 1, 2. So we can identify Vi with GF(5), and thus V = GF(5)2. Then {(1, 0), (2, 0), (3, 0), (4, 0)} and {(0, 1), (0, 2), (0, 3), (0, 5)}

are both orbits of D on V (as D~CD(V1) has an orbit of size 4 on Vi), and their union is an orbit of size 8 of G on V . Moreover, if a, b ∈ GF(5) − {0}, then (a, b)D

will contain (a, −b), since CD(V1) acts as x → −x on V2. Since D~CD(V1) has an or-

D bit of size 4 on V1, we also see that (a, b) will contain elements of the form (1, ∗), (2, ∗), (3, ∗), and (4, ∗). Altogether we see that S(a, b)DS = 8 = M. Putting this together shows that, since 8 = M, G has three orbits of size 8 on V , contradicting the hypothesis.

21 This concludes the case where D′ < G′. The Case where D′ = G′

We will consider the action of D~CD(V1) on V1 and CD(V1) acting on W1. Using Theorem 1, we have the following inequalities

′ SD ∶ D CD(V1)S ≤ M1 (6)

′ SCD(V1) ∶ CD(V1) S ≤ M2 (7)

where M1 is the largest orbit size of D~CD(V1) on V1 and M2 is the largest orbit

size of CD(V1) on W1. There are now four cases to consider; strict inequality in (6) and (7), strict inequality in (6) or (7) but not both, and equality in (6) and (7). First we consider the case where we have strict inequality in (6) and (7). Be-

′ ′ cause G is a p-group we know that pSD ∶ D CD(V1)S ≤ M1 and pSCD(V1) ∶ CD(V1) S ≤

M2. Therefore

2 ′ ′ M ≥ M1M2 ≥ p SD ∶ D CD(V1)SSCD(V1) ∶ CD(V1) S.

′ Recall SG ∶ DS = p and notice that SD ∶ D S ≤ MD ≤ pM1M2 so

2 ′ ′ ′ ′ ′ ′ p SD ∶ D CD(V1)SSCD(V1) ∶ CD(V1) S ≥ pSG ∶ DSSD ∶ D S = pSG ∶ D S = pSG ∶ G S > SG ∶ G S.

Putting the above equations together we have SG ∶ G′S < M. This contradicts our hypothesis that SG ∶ G′S = M, and therefore either (6) or (7) must be an equality.

′ Suppose that (6) is equal, that is SD ∶ D CD(V1)S = M1. If D~CD(V1) is abelian then by (4) we have D is abelian. If D is abelian then 1 = D′ = G′ and G

22 is abelian, a contradiction. Therefore we note that D~CD(V1) cannot be abelian for the rest of the paper. By Theorem 1 we have that D~CD(V1) has at least two orbits of size M1 on V1. Let v1, v2 ∈ V1 be representatives of two diﬀerent orbits of size M1 in the action of D~CD(V1) on V1. Let w ∈ W1 be in an orbit of size M2 in the action fo CD(V1) on W1.

′ Because (7) is strict, we have pSCD(V1) ∶ CD(V1) S ≤ M2. This gives us the following,

G D~CD(V1) CD(V1) M ≥ S(vi + w) S ≥ Svi SSw S = M1M2

′ ′ ′ ≥ pSD ∶ D CD(V1)SSCD(V1) ∶ CD(V1) S ≥ SG ∶ G S = M for i = 1, 2. This gives equality everywhere. By hypothesis G has exactly two orbits of size M on V . This means D~CD(V1) has exactly two orbits of size M1 on V1. By induction D~CD(V1) ≅ D8, SV1S = 9 and p = 2. Thus M1 = 4, and clearly M2 ≤ 4. This

′ gives us CD(V1) ≤ D8. If CD(V1) ≅ D8 then SCD(V1) ∶ CD(V1) S = 4 contradicting

(7) is strict. Therefore CD(V1) is size one, two, or four. This makes CD(V1) abelain.

′ That means M2 = SCD(V1)S = SCD(V1) ∶ CD(V1) S which contradicts (7) is strict. Therefore we know that (7) cannot be strict when (6) is equal.

′ We now consider the case that (6) and (7) are equalities. That is SD ∶ D CD(V1)S =

′ M1 and SCD(V1) ∶ CD(V1) S = M2. Then

′ ′ ′ ′ M = SG ∶ G S = pSDS~SD S = pSCD(V1) ∶ D CD(V1)SCD(V1) ∶ CD(V1) ∩ D S ≤ pM1M2 also,

M ≥ MD ≥ M1M2, so M1M2 ≤ MD ≤ M ≤ pM1M2. We know that exactly one of these inequalities is strict because SDS~SD′S < pSDS~SD′S = SGS~SG′S = M. We now have three cases to

23 consider.

In all of these cases we know that D~CD(V1) has at least two orbits of size p ′ ′ M1 on V1, otherwise D~CD(V1) would be abelian, making D < ( i=1D~CD(Vi)) = 1. Then we would have D′ = G′ = 1, contradicting that G is not abelian. Through- out the following arguments we will let v1, v2 ∈ V1 be representatives of two orbits of size M1 on V1.

We consider that M1M2 = MD = M < pM1M2. Assume that v3 ∈ V1 is a third orbit of size M1 in the action of D~CD(V1) on V1. Then let w1 ∈ W1 be a representative of an orbit of size M2 in the action of CD(V1) on W1. This gives

D D D us (v1 + w1) , (v2 + w1) , and (v3 + w1) ; three orbits of size MD = M on V . Thus we have three orbits are of size M in the action of G on V , a contradiction.

Therefore D~CD(V1) has exactly two orbits of size M1 on V1. Let w1, w2 ∈ W1 be representatives of distinct orbits of size M2 in the action of CD(V1) on W1. We

D D D D see (v1 + w1) , (v1 + w2) , (v2 + w1) , and CD(v2 + w2) are four orbits of size

MD = M on V , a contradiction. Therefore we know that CD(V1) has exactly one orbit of size M2 on W1. By Theorem 1 we see that CD(V1) is abelian. By induction we have that D~CD(V1) ≅ D8, p = 2, and V1 = V2 = V (2, 3). Therefore we have

CD(V1) × CD(V2) ⪅ D ⪅ D~CD(V1) × D~CD(V2) and CD(V1) ≤ D~CD(V1) ≅ D8.

If SCD(V1)S = 8, then CD(V1) ≅ D8, contradicting CD(V1) is abelian. If

SCD(V1)S = 4, then CD(V1) must be the Klein-4 which we have previously shown to be a contradiction. If SCD(V1)S = 2, then CD(V1) is Z2, the cyclic group of order two. If we examine the table above we see that all subgroups of D8 of order two have at least three orbits of size two in the action on V2. This is a contradiction. If

SCD(V1)S = 1, then D~CD(V1) has two orbits of size four in the action of D~CD(V1) on V1 and D~CD(V2) has two orbits of size four in the action of D~CD(V2) on V2. Therefore D has either four orbits of size four or an orbit of size eight. This contra-

24 dicts that G has exactly two orbits of size M = M1M2 = 4 in the action of G on V .

Therefore we know that M1M2 = MD = M cannot occur.

′ Suppose that M1M2 < MD = M = pM1M2. This gives us SDS~SD S < MD and we know D~CD(V1) has at least two orbits of size M1 in the action of D~CD(V1).

We claim that D~CD(V1) must have exactly two orbits of size M1 on V1. Let w ∈

D W1 be such that S(v1 + w) S = MD = M. Then any g ∈ G − D will stabilize the orbit

D g (v1 + w) . In particular, there is a g ∈ G − D such that (v1 + w) = v1 + w. Assume

D g d0 d0 d0 that g ∈ G − D also ﬁxes (v2 + w) ; then (v2 + w) = (v2 + w) = v2 + w for some d0 ∈ D. Recall that g ∈ G − D cycles around the elements of the components

g g V = V1 ⊕ ... ⊕ Vp. The ﬁrst component of (v1 + w) and (v2 + w) are therefore the

d0 d0 same, they are also v1 and v2 respectively. This means v1 = v2 contradicting v1, v2

D are from diﬀerent D~CD(V1)−orbits on V1. Hence g cannot ﬁx the orbit (v2 + w) . Therefore

G D M ≥ S(v2 + w) S ≥ pS(v2 + w) S ≥ pM1M2 = M.

G G Thus (v1 + w) and (v2 + w) are two orbits of size M in the action of G on V . We

D can then repeat this argument to show (v3 + w) must be a third distinct orbit of size MD = M, a contradiction. Therefore D~CD(V1) has exactly two orbits of size

M1 on V1. By induction we have D~CD(V1) ≅ D8, p = 2,V1 = V2 = V (2, 3). We also know CD(V1) × CD(V2) ⪅ D ⪅ D~CD(V1) × D~CD(V1) ≅ D8 × D8, and CD(V1) ≤ D8.

This tells us SCD(V1)S ∈ {1, 2, 4, 8}.

If SCD(V1)S = 8, then CD(V1) ≅ D8. This means D ≅ D8 × D8, SDS = 64, and

SGS = 128. By [10, Lemma 2.8] we have G < D8 ≀ Z2, where Z2. Because SD8 ≀ Z2S = 128 we have that G = D8 ≀ Z2, which is known to be not metabelian [4, Satz 15.3 (d)],

′′ ′ ′ ′ that is G ≠ 1. However we have that G = D = (D8 × D8) which is size four. This makes G′ abelian and G′′ = 1, a contradiction.

25 2 If SCD(V1)S = 4 we have that SDS < SD~CD(V1) × D~CD(V1)S = SDS ~16 = 64.

′ ′ ′ ′ That is SDS = 32. We also have D ≤ (D8 × D8) so SD S ≤ 4. Suppose SD S = 4 then

SDS 32 ′ SD′S = 4 = 8, so MD = 16 = M1M2 a contradiction. Suppose that SD S = 2. Notice

CD(V1) ∩ CD(V2) = 1 so CD(V1) × CD(V2) = 1 so CD(V1) × CD(V2) = CD(V1)CD(V2).

Let g ∈ G − D and (1, a) ∈ CD(V1) × CD(V2). Then

−1 −1 −1 g −1 ∗ ∗ −1 [(1, a), g] = (1, a) g (1, a)g = (1, a) (1, a) = (1, a )(a , 1) = (a , a )

∗ ′ ′ for some a ∈ CD(V1), and we have four choices for a ∈ CD(V1). Thus 2 = SD S = SG S ≥ 4, a contradiction.

2 SDS ′ If SCD(V1)S = 2, then 4 = 64 or SDS = 16. As before, we know that SD S ∈

′ SDS 16 {2, 4}. Suppose SD S = 4, then SD′S = 4 = 4, and MD = 8. We know (v1, 0) and (v2, 0)

are both in D-orbits of size four on V . Let w ∈ V2 be a regular orbit. Then (v1 + w)

′ would be in an orbit of size eight, contradicting M1M2 < MD. Suppose that SD S = 2, SDS then SD′S = 8 and MD = 16. Thus CD(V1) must have four orbits of size two on V2.

Let wi ∈ V2 for i = 1, 2, 3, 4 be representatives of these four orbits. Then we know by the table above that CD(V1) = {1, r}, because all other subgroups of D8 with size two have only three orbits of size two on V2. We also know there must exist a

d1 d2 d3 d4 di ∈ D, i = 1, 2, 3, 4 where w1 = w2, w2 = w3, w3 = w4, w4 = w1. Without loss let

d w1 = (1, 0) and w2 = (1, −1). Then there exists a d ∈ D where (1, 0) = (1, −1) in

′ the action of D on V2, a contradiction. If CD(V1) = 1 then D ≅ D8, but D8~D8 = 4 =

MD, a contradiction.

′ Suppose that SDS~SD S = M1M2 = MD < M. We know that D~CD(V1) has at

least two orbits of size M1 in the action of V1. Recall that v1 and v2 are representatives of these orbits.

Let us ﬁrst assume that CD(V1) has exactly one orbit of size M2 on W1.

26 By Theorem 1 CD(V1) is abelian. Therefore CD(Vi) is abelian for all i = 1, . . . , p,

and it follows that for each i, CD(Wi) is abelian. We further claim that for each i,

CD(Wi) has exactly one regular orbit of Vi. To see this, observe that CD(V1) has

exactly one regular orbit on W1. Hence CD(V1)~(CD(V1) ∩ CD(V2)) has exactly one

regular orbit on V2, and (CD(V1)∩CD(V2))~(CD(V1)∩CD(V2)∩CD(V3)) has exactly

one regular orbit on V3. We can repeate this until we ﬁnally get

p−1 p (∩j=1 CD(Vj))~(∩j=1CD(Vi)) ≅ CD(Wp)

has exactly one regular orbit on Vp. Since the actions of CD(Wi) on Vi are equiva- p lent for all i, the claim is true. Let A ∶= ∏i=1 CD(Wi). We see that A ⊴ G, and A = p i=1CD(Wi) is an internal direct product because CD(Wi)∩∏j∈{1,...,p}−{i} CD(Wj)) =

1, for i = 1, . . . , p (all elements in ∏j∈{1,...,p}−{i} CD(Wj)) act trivially on Vi). Apply- ing Lemma 2.2 to the action of CD(Wi) on Vi for all i. Putting this together thus shows that if we write VA = X1 ⊕...⊕Xm for some M ∈ N and irreducible A-modules such that V1 = X1 ⊕ ... ⊕ Xk for some k ∈ N, then m = kp and

m m SAS = M(SXiS − 1) = (SX1S − 1) ≤ M i=1

and k k M1 ≥ M(SXiS − 1) = (SX1S − 1) . i=1

Now recall that v1, v2 ∈ V1 are representatives of two orbits of size M1 of D~CD(V1)

on V1. Write vi = xi1 + ... + xik with xij ∈ Xj for j = 1, . . . , k, i = 1, 2. We may assume

that v1 is in a regular orbit of A~CA(V1), and thus x1j =~ 0 for j = 1, . . . , k. But then D A v1 = v1 = {y1 + ... + ykS0 =~ yi ∈ Xi for i = 1, . . . , k}, and this forces that x2j = 0 for at p−1 gj least one j ∈ {1, . . . , k}. If we let g ∈ G − D and put zi = vi + ∑ v1 for i = 1, 2, then j=1

27 it is clear that both z1 and z2 are in diﬀerent orbits of size greater than or equal to

SAS of G. Hence SG~G′S ≥ SAS.

s Now let q be the characteristic of V and write SX1S = q . Write SA~CA(X1)S = pt. Then pt = qs − 1 and hence with [10, Proposition 3.1] we know that either s = 1, p = 2, and q is a Fermat prime; or t = 1, q = 2, and p is a Mersenne prime; or s = 2, t = 3, p = 2, and q = 3. Moreover, by [10, Theorem 2.1] we know that

NG(X1)~CG(X1)Γ(X1) and since G is a p-group, altogether we conclude that

NG(X1)~CG(X1) ≅ A~CA(X1)

2 unless possibly in the third case, when SV1S = 9 and NG(X1)~CG(X1) ≅ Γ(3 ) is possible (in the ﬁrst case this is clear, in the second it follows by Fermat’s Little

Theorem). For the moment suppose that NG(X1)~CG(X1) ≅ A~CA(X1). Because

of the size of A with [10, Lemma 2.8] we conclude that G ≅ A~CA(X1) ≀ G~A where

G~A transitively and faithfully permutes the Xi (i = 1, . . . , m). Now with arguments similar to the one in the proof of [6, Lemma 2] we see that

m−1 m−1 S[A, G]S ≥ SA~CA(X1)S = (SX1S − 1) .

Moreover, by [2, Theorem 2.3] we have SG ∶ G′AS = SG~A ∶ (G~A)′S ≤ pm~p. Hence altogether we have

m ′ ′ ′ ′ (SX1S − 1) = SAS ≤ SG~G S = SG ∶ G ASSG A ∶ G S

m~p ′ ≤ p SA ∶ A ∩ G S

m~p ≤ p SA ∶ [A, G]S

m~p = p (SX1S − 1)

28 Now clearly SX1S − 1 ≥ p, and so it follows that

m m−1 m−1 p ≤ (SX1S − 1) ≤ p p .

So m − 1 ≤ m~p, and since m ≥ p, we get that p = m = 2, SX1S = 3, k = 1, V1 =

′ X1, SV S = 9, M = SG~G S = 4 and thus G is dihedral of order 8 acting on the nine elements of V . But then D is abelian, and since G′ = D′, G is also abelian. This is a contradiction.

In the exceptional case s = 2, t = 3, p = 2, q = 3 above we have that the kernel

m K~A of the permutation action of G~A on the Xi is of order at most 2 . So we see that G~Ω2(A) has A as an abelian normal subgroup, and so similarly as above

′ ′ ′ ′ SG~G S ≤ SG ∶ G KSSG K ∶ G S

m ′ ≤ 2 2 SK ∶ K ∩ G S

m ′ ≤ 2 2 SK~Ω1(A) ∶ (K ∩ G )Ω1(A)~Ω1(A)S ⋅ SΩ1(A)S

m−1 Now S[K~Ω1(A),G~Ω1(A)]S ≥ 4 , and thus altogether

3m m m ′ m m 2 = 8 = (SX1S − 1) = SAS ≤ SG~G S ≤ 2 2 SK~Ω1(A) ∶ [K~Ω1(A),G~Ω1(A)]S ⋅ 2 m m 8 m 2 2 2 ≤ ⋅ 4m−1 ⋅ 3 m+m−1 5 m+2 = 2 2 ⋅ 8 = 2 2

5 Hence 3m ≤ 2 m + 2 or, equivalently, m ≤ 4. Since m = kp = 2k, we have that m = 2 or m = 4.

If m = 2, then k = 1 and thus Xi = Vi for i = 1, 2. But then M1 = 8 = SV1S − 1,

and D~CD(V1) has exactly one orbit of size M1 on V1, contradicting our observation

29 above that D~CD(V1) has at least two orbits of size M1 on V1.

4 If m = 4, then k = 2 and SV1S = 3 . Hence G is isomorphic to a subgroup of

19 ′ m−1 GL(8, 3) and thus SGS ≤ 2 . As above, we know that SG S ≥ S[A, G]S ≥ (SX1S − 1) = 83 = 29, and hence SG~G′S ≤ 210 < 212 = SAS ≤ M, contradicting SGS~SG′S = M.

Therefore we know that CD(V1) has at least two orbits of size M2 on W1.

Let w1 and w2 be representatives of such orbits. If there exists a d ∈ CD(v1) such d that w1 = w2 we see that

′ ′ M ≥ MD ≥ M1pM2 ≥ pSD ∶ D S = SG~G ]

contradicting that MD < M. Therefore, no such d exists. This tells us that (v1 + w1) and (v1 + w2) lie in diﬀerent D-orbits on V . Similarly v2 + w1 and v2 + w1 are in diﬀerent D-orbits on V . p Now identify D with a subgroup of i=1D~CD(Vi). Also let g ∈ G − D and put p−1 p−1 gj D gj p Li = Q(vi ) ∶= {Q xjSxj ∈ (vi ) for j = 0, . . . , p − 1} ⊂ V j=0 j=0 for i = 1, 2. The Li are clearly G-invariant subsets, and L1 ∩ L2 = ∅. For any x ∈

D V2 ⊕ ... ⊕ Vp it follows that if the orbit (vi + x) ⊂ Li (i ∈ {1, 2}).

D We now have several cases to consider. The D-orbits (v1 + w1) and (v1 +

D D D w2) are both G−invariant. The D-orbits (v2 + w1) and (v2 + w2) are both G-

D D invariant. Lastly, one of the D-orbits (v1 + w1) or (v1 + w2) is not G-invariant,

D D and at least one of the orbits D-orbits (v2 + w1) or (v2 + w2) is not G-invariant.

D D Suppose that the D-orbits (v1 + w1) and (v1 + w2) are both G-invariant.

Then v1 + w1 ∈ L1 and v1 + w2 ∈ L1, and thus v2 + w1 ∉ L2 and v2 + w2 ∉ L2, that is

30 D D (v2 + w1) and (v2 + w2) are not G-invariant, so that

G ′ ′ M ≥ S(v2 + wi) S ≥ pS(v2 + wi)S ≥ pM1M2 ≥ pSD~D S = SG~G S

for i = 1, 2. This tells us that CD(V1) has exactly two orbits of size M2 in the action on W1.

Suppose that we have three orbits of size D~CD(V1) on V1, let v3 be a repre-

D D sentative of this orbit. We know that (v3 +w1) and (v3 +w2) are not G−invariant.

D D g That means (v2 + w1) = (v3 + w2) , so there exists a g ∈ G − D with (v1 + w2) = j g g g g g D v2 +w2 = v3 +w1 and w1 = v2 +x for some x ∈ W2. Therefore w1 ∈ L1 so v2 ∈ (v1 ) for

g g D g D g D some j. We can choose g so that j = 1 giving us that v2 ∈ (v1) but (v1) ∩ (v2) = ∅, a contradiction.

Therefore we have that D~CD(V1) has exactly two orbits of size M1 in the

action of V1. By induction D~CD(V1) ≅ D8, p = 2, and V1 = V (2, 3). We know that

CD(V1) × CD(V2) ≤ D ≤ D~CD(V1) × D~CD(V2) ≅ D8 × D8 and CD(V1) ≤ D~CD(V1) ≅

D8. Then SCD(V1)S ∈ {2, 4, 8}. If SCD(V1)S = 8 then D = D8 × D8, so SDS = 64,

′ SD S = 4 = SGS and SGS = 128. By [10, Lemma 2.8] we have G ≤ D8 ≀ Z2 is metabelian, but SG′S = 4 a contradiction.

If SCD(V1)S = 4, then CD(V1) ≅ Z4 (because CD(V1) has two orbits of size

M2 on V2, therefore it is not the Klein-4). Thus Z4 × Z4 ≤ D ≤ D8 × D8, SDS > 16,

′ ′ ′ and Z4 × Z4 < D so SDS = 32 and SD S = 2. Because D < (D8 × D8) we have that

′ ′ SDS 32 SD S ∈ {2, 4}. If SD S = 4 then SD′S = 4 = 8, contradicting that Z4 × Z4 has a regular

′ ′ orbit (size 16) on V. Therefore SD S = SG S = 2, but CD(V1) = 4 which we have shown makes SG′S ≥ 4, a contradiction.

If SCD(V1)S = 2, then CD(V1) ≅ Z2. This means that CD(V1) has at least three

orbits of size M2 on V2, a contradiction.

31 D D The case where (v2 +w1) and (v2 +w2) are both G-invariant will follow the same proof as in case 1 if we replace v1 with v2.

D D Suppose that at least one of the orbits (v1 + w1) or (v1 + w2) is not G-

D D invariant and at least one of the orbits (v2 + w1) and (v2 + w2) is not G-invariant.

D D Without loss we may assume that (v1 + w1) is not G-invariant. If (v2 + w1) is also

G G not G-invariant, we see that (v1 + w1) and (v2 + w1) are two distinct orbits of size M, because

G ′ M ≥ (vi + w1) ≥ pM1M2 = SG~G S for i = 1, 2 and if we write w1 = (x2, . . . , xp) ∈ V2 ⊕...⊕Vp then v1 +w1 = (v1, x2, . . . , vp) and v2 + w2 = (v2, x2, . . . , xp) have a diﬀerent number of components in the corre- sponding component of L1 and cannot be conjugate in G.

By induction we have that D~CD(V1) ≅ D8, V1 = V (2, 3), p = 2, and V =

V1 × V2. As shown above, this leads to a contradiction.

D D Hence (v2 + w1) is G-invariant and thus (v2 + w2) is not G-invariant. A

D similar argument then shows that (v1 + w2) must be G-invariant otherwise (v1 +

G G w2) and (v2 + w2) would be in two diﬀerent orbits of size M, and we could use induction as above to show D~CD(V1) ≅ D8, a contradiction.

D D D Thus (v1 + w1) and (v2 + w2) are both G-invariant. If (v1 + w2) and

D (v2 + w2) are G-conjugate then v2 + w1 ∈ L2 and v1 + w2 ∈ L1. This shows that v1 + w1 and v2 + w2 can only be conjugate in G if p = 2, so now let p = 2.

Suppose there are three orbits of size M2 in the action CD(V1) on V2. Let w3 ∈ W1 = V2 be a representative of such an orbit. As above it follows that v1 + w1, v1 + w2 and v1 + w3 all lie in diﬀerent D-orbits on V and so do v2 + w1, v2 + w2, and v2 + w3, and as above, with w3 in place of w2 we see the following version of case 3 must be true.

32 D D At least one of the orbits (v1 + w1) or (v1 + w3) is not G-invariant, and

D D at least one of the orbits (v2 + w1) or (v2 + w3) is not G-invariant. We already

D D know that (v2 + w1) is G-invariant, it follows that (v2 + w3) is not G-invariant.

G G The argument from an earlier shows (v2 + w2) and (v2 + w3) are diﬀerent G- orbits. Assume that there is third orbit of size M1 in the action of D~CD(V1) on

V1, let v3 be a representative of this orbit. Then at least two or the three orbits

D D D D (v3+w1) , (v3+w2) , and (v3+w3) are not G-invariant. If (v3+w1) is G-invariant,

G G G then (v3 + w2) , (v3 + w1) , and (v2 + w2) are three G-orbits of size M, a contra-

D G G G diction. If (v3 + w2) is G-invariant, then (v2 + w3) , (v2 + w2) , and (v3, w3) are

D three G-orbits of size M, a contradiction. Lastly, if (v3 + w3) is G-invariant, the

G G G (v3 + w1) , (v2 + w2) , and (v3 + w2) are three G-orbits of size M, a contradiction.

We see that D~CD(V1) has exactly two orbits of size M1 on V1. By induction we have that D~CD(V1) ≅ D8, a contradiction. This concludes equality in (6) and (7). Suppose we have equality in (7) and strict inequality in (6). That is

′ ′ ′ ′ M ≥ M1M2 ≥ pSD ∶ D CD(V1)SSCD(V1) ∶ CD(V1) S ≥ pSD~D S = SG ∶ G S.

′ Because SG ∶ G S = M we have equality everywhere, and M = M1M2,M1 > SD ∶

′ ′ D CD(V1)S,M2 = SCD(V1) ∶ CD(V1) S. Again let MD denote the largest orbit size of

D on V , then MD ≥ M1M2 so MD = M. By Theorem 1 CD(V1) is abelian or has at least two orbits of size M2 of D on W1.

Suppose CD(V1) has at least two orbits of size M2 on W1, and let w1, w2 ∈

W1 be representatives of these orbits. Let D~CD(V1) have at least two orbits of size

D D D M1 on V1. Because M = M1M2 we have that (v1 + w1) , (v1 + w2) , and (v2 + w2)

D and (v2 + w1) are all (not necessarily distinct) orbits of size MD = M. This means

D D that (v1 + w1) is G-invariant, meaning that (v2 + w1) is not G-invariant. That is

33 G S(v2 + w1) S > M1M2 = M, a contradiction.

Therefore we have that D~CD(V1) has exactly one orbit of size M1 on V1. If

G (v1 + w) is not G−invariant then S(v1 + w) S = M, but D has two orbit of size M on

D V , diﬀerent from (v1 + w) giving G three G-orbits of size M on V , a contradiction.

D Therefore (v1 + w) is G-invariant. If CD(V1) has two orbits of size M2 on W1, then

D D (v + w1) and (v + w2) are two D−orbits of size M. Thus CD(V1) has exactly two

orbits of size M2 on W1. If M2 > M1, then CD(V1) < D would give us an orbit of

size M2 on V1 contradicting that M1 was the size of the largest orbit of D on V1. If

M2 = M1 then CD(V1) would have two orbits of size M1 on V1 contradicting that

D has exactly one orbit of size M1 on V1. If p = 2 then M1 > M2. Let g ∈ G − D

which permutes the elements of the components V1 → V2 → ... → Vp → V1. Let

g g D v = x ∈ V2 and w1 = y ∈ W2. We claim that (x + y) ∉ (v + w1) . Suppose not, then

D D (x+y) = (v +w1) . The V2 component of the left hand side has M1 choices for this

element. However, the right hand side has M2 choices for elements in V2 Therefore

(x + y)D is a third orbit of size M in the action of D on V , a contradiction. This means p ≠ 2. Using [10][Theorem 4.4] we know that p would have to be a Mersenne prime,

contradicting the original hypothesis that SGS is not divisible by a Mersenne prime.

Suppose that CD(V1) has exactly one orbit of size M2 on W1, then CD(V1) is abelian by Theorem 1. As argued above and using Lemma 1 we conclude that there

is an irreducible normal subgroup of G such that if we write VA = X1 ⊕ ... ⊕ Xm for

m some m ∈ N and irreducible A-modules Xi, then SAS = (SXiS − 1) ≤ M. Without

loss we may assume that V1 = X1 ⊕ ... ⊕ Xk for some k ∈ N (which divides m), and

k then M1 ≥ (SX1S − 1) = SA~CA(V1)S. Now CA(V1) ≅ A~CA(W1) is isomorphic to

a maximal abelian subgroup of D~CD(V1). If there was a larger abelian subgroup,

call in V , this subgroup would still act on each Xj, (j = k + 1, . . . , m) where for

34 Yi = `∈{1,...,m}−{i}X` we put

m Cj = Xi∈{1,...,m}−{j}CA(Yi) ≤ A = i=1CA(Yi).

Hence there would exist a j ∈ {k + 1, . . . , m} such that SBCB(Xj)S > SA~CA(Xj)S =

SXjS−1, a contradiction. Hence CA(V1) ≅ A~CA(W1) is self centralizing in D~CD(W1).

But as D~CD(V1) is nonabelian, clearly D~CD(W1) is nonabelian, and thus there

′ exists a d ∈ D and a ∈ CD(V1) such that 1 ≠ [d, a]. Hence D ∩ CA(V1) > 1. On the

′ ′ ′ other hand, we have, as CD(V1) = 1 and G = D , that

′ ′ M = pM1SCD(V1)S = SG~G S = pSD~D S

′ ′ ′ = pSD ∶ D CD(V1)SSD CD(V1) ∶ D S

′ ′ = pM1SD CD(V1) ∶ D S

′ = pSM1CD(V1) ∶ (D ∩ CD(V1))S.

′ This forces D ∩ CD(V1) = 1, and we have a contradiction. This concludes our proof.

35 IV. The Open Case

In this section we will reiterate the conditions under which we use the Mersenne prime condition. Let the action of G on V be imprimitive where D′ = G′. Let

M1 be the size of the largest orbit of D~CD(V1) on V1 and M2 be the largest or-

′ ′ bit size of CD(V1) on M2. If SD ∶ D CD(V1)S < M1 , SCD(V1) ∶ CD(V1) S = M2 and

′ ′ M1M2 = MD = M = pSDS~SD S = SGS~SG S, D~CD(V1) has exactly one orbit of size M1 on V1 and CD(V1) has exactly two orbits of size M2 on W1 then we know that p is a Mersenne and q = 2. However we are almost certain this case cannot happen.

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